Question

If $\tan A=\frac{1}{7}$ and $\tan B=\frac{1}{3}$, show that cos 2A = sin 4B

Answer 1

Given:
$\tan A=\frac{1}{7}$ and $\tan B=\frac{1}{3}$

Using the identity $\tan 2B=\frac{2\tan B}{1-{\tan }^{2}B}$, we get

$\tan 2B=\frac{2\times {\displaystyle \frac{1}{3}}}{1-{\displaystyle \frac{1}{9}}}=\frac{3}{4}$
Now, using the identities $\cos 2A=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}\mathrm{and}\sin 4B=\frac{2\tan 2B}{1+{\tan }^{2}2B}$, we get

$$\begin{gathered} \cos 2A=\frac{1-{\left({\displaystyle \frac{1}{7}}\right)}^2}{1+{\left({\displaystyle \frac{1}{7}}\right)}^2}\mathrm{and}\sin 4B=\frac{2\times {\displaystyle \frac{3}{4}}}{1+{\left({\displaystyle \frac{3}{4}}\right)}^2} \\ \Rightarrow \cos 2A=\frac{48}{50}\mathrm{and}\sin 4B=\frac{2\times {\displaystyle \frac{3}{4}\times 16}}{25} \\ \Rightarrow \cos 2A=\frac{24}{25}\mathrm{and}\sin 4B=\frac{{\displaystyle 24}}{25} \end{gathered}$$

∴ cos 2A = sin 4B